Triangle XYZ is equilateral. Points Y and Z lie on a circle centered at O such that X is the circumcenter of triangle OYZ and X lies inside triangle OYZ. If the area of the circle is 48pi, then find the area of triangle XYZ.
Please help I am very stuck on this problem and unsure how to do it.
Let the radius of the circle be r. Since X is the circumcenter of triangle OYZ, it lies on the perpendicular bisectors of all three sides of the triangle. This means that XY = XZ = R, and YZ = 2R/sqrt(3), which is the diameter of the circle.
The area of triangle XYZ is given by
A = r * s / 2
where s is the semiperimeter of the triangle. In this case, s = R + R = 2R.
Plugging in the area of the circle, we get
A = 48pi * 2R / 2 = 96piR / 2
Since R = sqrt(48pi), we can simplify the expression to get
A = 96pi * sqrt(48pi) / 2 = 24sqrt(3)pi
Therefore, the area of triangle XYZ is 24sqrt(3)pi.